Adjoint Transformations and Self-Adjoint Operators I don't quite understand the whole adjoint and self adjoint thing. I know their definitions:

Given a linear transformation $A:\mathbb{R}^d \to \mathbb{R}^m$, its adjoint >transformation, $A^*:\mathbb{R}^m \to \mathbb{R}^d$ is defined by $$(Ax,y)=(x,A^*y).$$
A linear operator $B:\mathbb{R}^k \to \mathbb{R}^k$ is self-adjoint if $B=B^*$.

So, $B$ is defined by $(Bx,y)=(x,By)$, right? Also, could someone please give me some examples of both $A$ and $B$? Thanks in advance.
 A: We can't say that any $B$ is defined by $(Bx,y) = (x,By)$.  However, if $B$ is self-adjoint, then $B$ has the property that $(Bx,y) = (x,By)$ for every $x,y \in \Bbb R^k$.
Here are some examples with matrices: recall that the usual inner product for $x,y \in \Bbb R^n$ is given by
$$
(x,y) = x^Ty
$$

Define the transformation $T_A: \Bbb R^3 \to \Bbb R$ by $A = \pmatrix{1&2&3}$.  That is,
$$
T_A(x) = Ax = x_1 + 2x_2 + 3x_3
$$
The adjoint $T_A^*$ (with matrix $A^*$) is defined by the property
$$
(Ax,y) = (x,A^*y)
$$
for all $x \in \Bbb R^3$ and $y \in \Bbb R$.  Since $A^* \in \Bbb R^{3 \times 1}$, it suffices to compute $T_{A^*}(1) = A^*1 = A^*$.  We note that for all $x$, we have
$$
x^TA^* = (x,A^*1) = (Ax,1) = x_1 + 2x_2 + 3x_3
$$
So, in particular, we must have
$$
A^* = \pmatrix{1\\2\\3}
$$

Define the transformation $T_B:\Bbb R^2 \to \Bbb R^2$ by the matrix
$$
B = \pmatrix{0&1\\1&2}
$$
For any $x,y \in \Bbb R^2$, we compute
$$
(Bx,y) = \left( \pmatrix{x_2\\x_1 + 2x_2}, \pmatrix{y_1\\y_2} \right)
= x_2 y_1 + (x_1 + 2x_2)y_2 = x_1 y_2 + x_2y_1 + 2x_2 y_2
$$
Similarly,
$$
(x,By) = \left( \pmatrix{x_1 \\ x_2}, \pmatrix{y_2\\y_1 + 2y_2} \right)
= x_1 y_2 + x_2(y_1 + 2y_2) = x_1 y_2 + x_2y_1 + 2x_2 y_2
$$
So, we say that $B$ is self-adjoint.
